BMAD Posted August 25, 2015 Report Share Posted August 25, 2015 What would be the remainder when 3^4^5^6^7 ... so on through infinity is divided by 17? Quote Link to comment Share on other sites More sharing options...

0 araver Posted August 25, 2015 Report Share Posted August 25, 2015 Assuming exponentiation. First: 3^4^(2k+1) = 13 (mod 17) for any positive integer k.Fermat's little theorem states that for p = 17 (prime) a^17 = a (mod 17), which for a = 3 with gcd(3,17)=1 means 3^16=1 (mod 17).So 3^4^(2k+1) = (3^4^2)^k * 3^4 = 1*3^4 = 13 (mod 17). I don't really like infinities but for any member in the series x_n = 3^4^5^.. ^n, (with n>=5) there exists a positive integer m such that 5^.. ^n = 2*m+1 (as it is odd). Therefore x_n = 13 (mod 17) using k=m in the lemma above.So, the answer to the question would be 13 i.e. the series y_n = x_n (mod 17) converges to 13 as n goes to infinity. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted August 25, 2015 Report Share Posted August 25, 2015 I was going to ask which direction the exponentials are evaluated. But obviously 3 is first raised to the 4th power; that result is then raised to the 5th power; and so on. Only finite power towers can be evaluated right to left.Unless you are using ^ to denote multiplication. In which the remainder is zero. Quote Link to comment Share on other sites More sharing options...

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